Question

When two cells of emfs. E1 and E2 are connected in series so as to assist each other, their balancing length on a potentiometer is found to be 2.7 m. When the cells are connected in series so as to oppose each other, the balancing length is found to be 0.3 m. Compare the emfs of the two cells.

Answer 1

Data : l₁ = 2.7m (cells assisting),
l₂ = 0.3 m (cells opposing)
E₁ + E₂ = Kl₁ and E₁ – E₂ = Kl₂
∴\(\frac{E_{1}+E_{2}}{E_{1}-E_{2}}=\frac{K l_{1}}{K l_{2}}\)
∴ \(\frac{E_{1}}{E_{2}}=\frac{l_{1}+l_{2}}{l_{1}-l_{2}}=\frac{2.7+0.3}{2.7-0.3}=\frac{3}{2.4}=\frac{30}{24}\) = 1.25
The ratio of the emf’s of the two cells is 1.25.